查看: 51137|回復: 46

[SQL] 這幾年牛蛙寫過的奇葩代碼

[復制鏈接]
論壇徽章:
79
生肖徽章2007版:牛
日期:2012-08-02 22:43:00紫蛋頭
日期:2012-12-08 09:43:38鮮花蛋
日期:2012-11-17 12:02:07鮮花蛋
日期:2013-02-05 21:53:34復活蛋
日期:2012-11-17 12:02:07SQL極客
日期:2013-12-09 14:13:35SQL數據庫編程大師
日期:2013-12-06 13:59:43SQL大賽參與紀念
日期:2013-12-06 14:10:50ITPUB季度 技術新星
日期:2012-11-27 10:16:10最佳人氣徽章
日期:2013-03-19 17:24:25
跳轉到指定樓層
1#
發表于 2016-8-31 23:55 | 只看該作者 回帖獎勵 |倒序瀏覽 |閱讀模式
本帖最后由 udfrog 于 2016-9-1 17:24 編輯

之前自己整理過一個文檔,存著自己這幾年在pub上寫過的有趣代碼。今天又追加了幾個今年寫的,一并分享出來,如果有緣人喜歡,那就很好。
我這個人不是個著調的開發,從來不愛寫注釋,所以看著痛苦的話,抱歉了。格式也不夠好。
這幾年寫下來,我也總在體會,我自認為我的sql寫得很不同尋常,的好,最根本的原因就是我對問題建模所產生的數據結構有相當的優越性,那優秀的算法自然也就不難形成。


巧的是,論壇新開了打賞功能,我覺得挺有意思的,如果喜歡,可以盡情打賞,讓我嘗個鮮。


1. Adjacent Digits


How many positive 10-digit integers are there with the property that for each digit N, the number of adjacent N's is at most N?


(Examples: 1223334444, 7777777822, 5999999999, 3434343434)


這個沒去優化,在我的pc上大概要跑8分鐘,也不清楚是否可以優化
with b as (
select level lv from dual connect by level<=9),
t1 (str) as (
select rpad(lv, level, lv)
from    b
connect by level<=lv and prior lv=lv and prior dbms_random.value()>0
),
t as (
select * from t1 where length(str)<=5
),
r (str) as (
select  str
from    t
union all
select  r.str||t.str
from    t, r
where   substr(r.str, -1, 1)!=substr(t.str, 1, 1)
and     length(r.str||t.str)<=5
),
r2 as (
select  str, substr(str, -1) trail, 5-nvl(length(rtrim(str, substr(str, -1))), 0) len
from    r
where   length(str)=5
),
r3 as (
select  str, substr(str, 1, 1) lead, 5-nvl(length(ltrim(str, substr(str, 1, 1))), 0) len
from    r
where   length(str)=5
)
select  56143*56143-count(*)
from    r2, r3
where   r2.trail=r3.lead
and     r2.len+r3.len>r2.trail;




2.Double Factored
Let''s call a number "double factored" if at least one of its prime factors repeats itself when factorised. What is the smallest positive integer that itself and its four neighbors (2 before, 1 before, the number itself, 1 after, 2 after) are double factored?
If the problem was asked for the number and its two neighbors (1 before, the number itself, 1 after) then the answer would be 49. (48=2x2x2x2x3 , 49=7x7, 50=2x5x5).


with df as (
select (rownum+1)*(rownum+1) df from dual connect by rownum<=20),
t as (
select rownum f from dual connect by rownum<=1000),
tmp as (
select  min(df*f) keep (dense_rank first order by df*f) r, row_number() over (order by df*f) rn
from  df, t
group by df*f)
select  min(a.r)-2
from  tmp a, tmp b
where a.r=b.r+4
and a.rn=b.rn+4;




3.Full Prime Number


All the numbers that can be read with two or more adjacent digits inside a prime number are also prime numbers. What is the greatest number having this property?


Exampe:6173 is such a number. Because 61, 17, 73, 617, 173 and 6173 are prime numbers


WITH t0 AS (
    SELECT 2*ROWNUM+1 rn FROM DUAL CONNECT BY ROWNUM <= (100000)/2-1-1
    ),
t as(SELECT rn from t0
      where mod(rn,3)<>0
      and mod(rn,5)<>0
      and mod(rn,7)<>0
    )
,p AS (
SELECT TO_CHAR(rn) n,LENGTH(rn) len
   FROM (SELECT rn from t
         MINUS
         SELECT t1.rn * t2.rn
           FROM t t1, t t2
         WHERE t1.rn <= t2.rn
            AND t1.rn BETWEEN 11 AND SQRT(100000)
            AND t1.rn * t2.rn <100000
       )
WHERE rn>10
),
tmp (n, digit) as (
select  n, 2 digit
from  p
where len=2
union all
select  a.n||substr(b.n, -1) n, a.digit+1 digit
from  tmp a, tmp b, p
where a.digit=b.digit
and substr(a.n, 2)=substr(b.n, 1, b.digit-1)
and a.n||substr(b.n, -1)=p.n)
select max(to_number(n)) from tmp;




4.Prime Sums


All of the digits of a number are different. The sums of all neighboring three digits of this number are prime numbers. What is the largest number having these properties?


with b as (
select rownum-1 n from dual connect by rownum<=10),
p as (
select level n from dual where (level>=2 and mod(level, 2)<>0 and mod(level, 3)<>0 and mod(level, 5)<>0) or level in (2, 3, 5) connect by level<=23),
tmp as (
select     d1.n||d2.n||d3.n d
from       b d1, b d2, b d3, p
where      d1.n<>d2.n
and        d1.n<>d3.n
and        d2.n<>d3.n
and        d1.n+d2.n+d3.n=p.n),
t (d) as (
select     d
from       tmp
where      not d like '0%'
union all
select     t.d||substr(tmp.d, -1)
from       t, tmp
where      substr(t.d, -2)=substr(tmp.d, 1, 2)
and        instr(t.d, substr(tmp.d, -1))=0)
select max(to_number(d)) from t;




5.Test


35 students took a test of 100 problems.


-Each problem is solved by exactly one girl and one boy.
-There are at least one girl who solved exactly one problem, and at least one girl who solved exactly two problems.
-There are at least one boy who solved exactly four problems, and at least one boy who solved exactly five problems.
-The number of the problems solved by the girl who has solved the maximum number of problems is X.
-The number of the problems solved by the boy who has solved the maximum number of problems is Y.


If the minimum values for both X and Y are the same, find the number of girls who have taken the test.


select  n
from  (select rownum+2 n from dual connect by rownum<=35-2-2-1)
where ceil((100-9)/(35-n-2))=ceil((100-3)/(n-2));




6.Solve Problem: Complex Sequence Generation
One of the questions on the Roundtable discussion was: "On one of the forums (SQL.ru) sometimes one of the members challenge everybody by posting a task (Friday challenge) and seeking for the best solution. I remember we had something like that about year ago here as well. Can we bring that back?"
Our "Solve Problem" feature never went away - we just hadn't posted one for a while. And Faig's question spurred me to take action. So we now have up on the website the following coding challenge:
Oracle sequences generate the next integer value available in the sequence. But sometimes applications rely on keys with a more complex structure. In this challenge, a player from Islamabad asks for some help to generate an alpha-numeric value in a sequence that follows a very interesting set of rules (explained in the question). He writes:
I need to write a function with this header:
FUNCTION plch_next_key (from_this_key_in in varchar2) RETURN VARCHAR2in which the key is a 5 digit alpha-numeric sequence of values, each of which has two parts:


1. Letter part: letters from A-Z, a-z. The number of letters increase from left to right
2. Digit part: left filled with 0s to length 5, after letters.
The rightmost letter stays fixed, and the remainder (a number starting with 1) is incremented, is then left padded with 0s to length 5 (including letters), up to all 9s.
When you have all 9s and the rightmost letter is not z, then increment the rightmost letter through the sequence A-Z, a-z and the remaining digits are reset to 1 (padded left with 0s)s.
When the rightmost letter is z, but there are non-z letters to the left, the rightmost z and the other letters are incremented A-Z, a-z as needed, with the rightmost z's becoming A's and the 9s are reset to 1 (padded left with 0s).
When all letters are z and all digits are 9s, then replace each z with A, replace the first 9 with A, and the remaining digits are reset to 1 (padded left with 0s).
Here's a partial set of key ranges to illustrate these rules:
A0001...A9999B0001...B9999....Z0001...Z9999a0001...z9999AA001...AA999AB001...AB999....AZ001...AZ999Aa001...Az999BA001...BZ999Ba001...Bz999CA001...zz999AAA01...zzz99AAAA1...zzzz9AAAAA...zzzzzYour mission, if you choose to accept it, is to implement this function so that it performs efficiently and is easy to understand.


create or replace function plch_next_key(from_this_key_in in varchar2)
return varchar2
is
len int:=5;
v_digit varchar2(32);
v_char varchar2(32);
begin
v_char:=regexp_substr(from_this_key_in, '\D+');
v_digit:=regexp_substr(from_this_key_in, '\d+');
return case when replace(v_digit, '9') is not null then
v_char||lpad((v_digit+1), length(v_digit),'0')
else
substr(
regexp_substr(v_char, '(.*?)(.??)(z*)$', 1, 1, 'c', 1)||
nvl(replace(chr(ascii(regexp_substr(v_char, '(.*?)(.??)(z*)$', 1, 1, 'c', 2))+1), '[', 'a'), 'A')||
replace(regexp_substr(v_char, '(.*?)(.??)(z*)$', 1, 1, 'c', 3), 'z', 'A')||
rpad('0', length(v_digit)+nvl(sign(length(replace(v_char, 'z'))), 0)-2, '0')||'1',
1, len
)
end;
end;
/




7.Four Digits


You have a number where any digit appears at most twice. The sum of all neighboring four digits in this number is a square number.


What is the maximum possible value for this number?


Example: 205290 is such a number because no digit appears more than twice and 2+0+5+2, 0+5+2+9, and 5+2+9+0 are square numbers.




with t as (
  select to_char(level-1) l from dual connect by level<=10),
t2 as(
select
a.l||b.l||c.l||d.l v4 --??
from t a,t b,t c,t d
where a.l+b.l+c.l+d.l in(4,9,16,25)
and a.l<>0
),
r(vx,lv)as(
select cast(v4 as varchar(40)),1 from t2
union all
select vx||l,lv+1 from r,t
where
instr(vx,l,1,2)=0 --????2?,????????????
and substr(vx,-1,1)+substr(vx,-2,1)+substr(vx,-3,1)+l in(4,9,16,25)
and lv<=16)
select to_char(max(0+vx)) from r where regexp_substr(vx, '([[:digit:]])(\1){2}') is not null;




8.Ten Numbers


_
|_|_ _
|_|_|_|
|_|_|_|
|_|_|_|


Place each of the ten numbers between 1 and 10 in the squares so that no two adjacent squares (horizontally or vertically) should contain;


consecutive numbers,
numbers with the same parity (odd or even).


In how many different ways can this task be accomplished?


with t as (
select rownum n from dual connect by rownum<=8)
select count(*)*2
from t
where  level=8 and abs(connect_by_root(n)-n) in (3, 5, 7)
start with n in (2, 4, 6)
connect by nocycle abs(prior n -n) in (3, 5, 7);




9.Plastic Digits


There are four sets of plastic digits. Each set has four digits (1, 2, 3, 4) and each set has a different colour (red, blue, green, yellow). You will place these 16 digits into 4x4 table such that every two adjacent squares (vertically or horizontally) should have either the same color or the same number.>

In how many different ways can this placement be done?>

If the problem was asked for 2 digits, 2 colors and a 2x2 table, then the answer would be 8.
  
  
WITH d AS (
SELECT n,c,POWER(2,ROWNUM-1) AS id
   FROM (SELECT LEVEL n FROM DUAL CONNECT BY LEVEL<=4)
       ,(SELECT 'R' c FROM DUAL UNION ALL SELECT 'B' c FROM DUAL UNION ALL SELECT 'G' c FROM DUAL UNION ALL SELECT 'Y' c FROM DUAL)
)
,t(cnt,bits,n0,c0,n1,c1,n2,c2,n3,c3) AS (
SELECT 1
       ,id
       ,null
      ,null
       ,null
      ,null
       ,null
      ,null
       ,n
       ,c
   FROM d
where        rownum=1
UNION ALL
SELECT t.cnt+1
       ,t.bits+d.id
       ,nvl(t.n1, d.n)
       ,nvl(t.c1, d.c)
       ,t.n2
       ,t.c2
       ,t.n3
       ,t.c3
       ,d.n
       ,d.c
FROM t,d
WHERE cnt<16
        AND BITAND(t.bits,d.id)=0
        AND (MOD(t.cnt,4)=0 OR d.n=t.n3 OR d.c=t.c3)
        AND (t.cnt<4 OR d.n = t.n0 OR d.c = t.c0)
)
SELECT COUNT(*)*16 FROM t WHERE cnt=16;




10.Balls And Boxes

What is the number of different ways to distribute different colored ten balls into four different boxes?

-All boxes will have different number of balls.
-An empty box is allowed.

If the problem was asked for 3 balls (white, black, red) and 2 boxes, then the answer would be 8.
1. (empty) - (w,b,r), 2. (w,b,r) - (empty), 3. (w) - (b,r), 4. (b) - (w,r), 5. (r) - (b,w), 6. (b,r) - (w)
7. (w,r) - (b), 8. (b,w) - (r)


此解法刻意追求代碼的趣味性,性能遠非最優
with t as (
select  sign(bitand((level-1), 1))+sign(bitand((level-1), 2))+sign(bitand((level-1), 4))+sign(bitand((level-1), 8))+
  sign(bitand((level-1), 16))+sign(bitand((level-1), 32))+sign(bitand((level-1), 64))+sign(bitand((level-1), 128))+
  sign(bitand((level-1), 256))+sign(bitand((level-1), 512)) s, rownum-1 n from dual connect by level<=1024)
select  count(*)*24
from  t t1, t t2, t t3, t t4
where bitand(t1.n, t2.n)=0
and bitand(t1.n, t3.n)=0
and bitand(t1.n, t4.n)=0
and bitand(t2.n, t3.n)=0
and bitand(t2.n, t4.n)=0
and bitand(t3.n, t4.n)=0
and t1.n+t2.n+t3.n+t4.n=1023
and t1.s>t2.s
and t2.s>t3.s
and t3.s>t4.s;




11.Two Pawns

Five pieces are randomly selected from a standard chess set. If it is known that at least one of these pieces is a pawn, what is the probability that at least two of these five pieces are white pawns?

Note:In a standard chess set there are 8 pawns, 2 rooks, 2 bishops, 2 knights, 1 queen, and 1 king for each color (white and black) making a total of 32 pieces.

After simplification, enter your answer as a/b.


with t as (
select case when rownum<=8 then 0 when rownum<=16 then 1 else 2 end s, rownum n from dual connect by rownum<=32)
select  count(case when regexp_count(sys_connect_by_path(s, ','), '0')>=2 then 1 else null end)||'/'||count(*)
from  t
where level=5
connect by prior n<n and level<=5
start with s in (0, 1);




12.Digits

A number has different digits, and the digits of the sum of each neighboring two digits are not used in this number.

What is the largest number satisfying these conditions?

Example:
7692
7+6=13, 6+9=15, 9+2=11 (The digits 1, 3 and 5 are not used.)


with t as (
select to_char(rownum-1) n from dual connect by rownum<=10),
r(n, p, na) as (
select     n, n, n
from       t
union all
select     t.n, r.p||t.n, r.na||(r.p+t.n)
from       r, t
where      translate(r.p||t.n, (r.n+t.n)||'x', 'x')=r.p||t.n
and instr(r.p, t.n)=0
and instr(r.na, t.n)=0)
select     max(to_number(p)) from r;




13.        千王之王
52張牌,四張A,隨機打亂后問,從左到右一張一張翻直到出現第一張A,請問平均要翻幾張牌?


with b as (
select rownum n from dual connect by rownum<=51),
f (fac, n, lv) as (
select b.n, b.n, b.n from b where n>=2
union all
select     f.fac*b.n, f.n, f.lv-1
from       f, b
where      f.lv-1=b.n),
p (p, n, lv) as (
select     48, 48, 1
from       dual
union all
select     p.p*b.n, p.n-1, p.lv+1
from       p, b
where      p.n-1=b.n)
select     sum(b.n*nvl(p.p, 1)*f.fac)/sum(nvl(p.p, 1)*f.fac)
from       f, p, b
where      f.lv=1
and (52-f.n)=b.n
and b.n-1=p.lv(+)
and b.n<=49;




14.        構造最大數
給定只包含正數的數組,給出一個方法,將數組中的數拼接起來,得到的數,是最大的。


with a  as (
select 4 x from dual union all
select 94 x from dual union all
select 14 x from dual union all
select 9 x from dual union all
select 1 x from dual),
b as (
select  max(length(x)) len
from    a)
select  listagg(x, '') within group (order by rpad(x, len*2, x) desc)
from    a, b;


注:關于此算法rpad(x, len*2, x)的證明,在101樓
http://www.3490925.live/forum.php?mod=viewthread&tid=1804279&page=11#pid21599534




15.        編碼任務
假設有一個池塘,里面有無窮多的水,F有2個空水壺,容積分別為5升和6升。如何只用這2個水壺從池塘里取得3升的水(最后,這三升水,在其中一個壺里)


with
t(n1, n2, n3, n4, lv) as (
select 5, 0, 6, 0, 1
from DUAL
union all
select decode(t.n1, 0, 5, decode(t.n2, 6, t.n1, greatest(t.n1-(6-t.n2), 0))) n1,
decode(t.n1, 0, t.n2, decode(t.n2, 6, 0, least(t.n1+t.n2, 6))) n2,
decode(t.n3, 0, 6, decode(t.n4, 5, t.n3, greatest(t.n3-(5-t.n4), 0))) n3,
decode(t.n3, 0, t.n4, decode(t.n4, 5, 0, least(t.n3+t.n4, 5))) n4,
t.lv+1
from t
where --t.lv+1 <=42
t.n1!=3
and t.n2!=3
and t.n3!=3
and t.n4!=3
)
select *
from t;




16. 三只青蛙的換位游戲
百度一個可用的鏈接 http://www.3366.com/flash/32099.shtml


with t as (
select round(2.5-level) pos, to_char(ceil(level/2)) c
from dual
  connect by level<=4
),
r (s, x_pos, path) as (
  select '222x111' s, 4, ''
  from dual
  union all
  select substr(replace(r.s, 'x', t.c), 1, x_pos+t.pos-1)||'x'||substr(replace(r.s, 'x', t.c), x_pos+t.pos+1), x_pos+t.pos, path||to_char(x_pos+t.pos)||'#'
  from t, r
  where substr(r.s, x_pos+t.pos, 1)=t.c
  and x_pos+t.pos between 1 and 7
)
select *
from r
where s='111x222';




17. 第三屆sql大賽第一題
題目鏈接 http://www.3490925.live/thread-1943911-1-1.html


相當得意的一段代碼
insert into TICTACTOE
with b as (
select level||'' n, decode(level, 1,111,2,1001,3,10100001,4,10010,5,111100,6,10010000,7,1100010,8,1001000,11000100) v from dual connect by level<=9
),
r(p, xp, op, i, w) as (
select n, v, 0, 0, ''
from b
union all
select p||n, decode(i, 1, v, 0)+xp, decode(i, 1, 0, v)+op, mod(i+1, 2),
case when instr(decode(i, 1, v, 0)+xp, '3')>0 then 'X' else case when instr(decode(i, 1, 0, v)+op, '3')>0 then 'O' else null end end
from r, b
where instr(r.p, n)=0
and w is null
)
select p, translate(translate('123456789', p, 'XOXOXOXOX'), '123456789', '---------'), nvl(w, 'D')
from r
where w is not null or length(p)=9;




18. 12個人帶7種顏色帽子,出現不少于3個相同的概率
qingyun搞來的一個題目 http://www.3490925.live/thread-1905121-1-1.html


直接貼newkid不厭其煩幫我改兩步來加速的版本
with t as (
select  power(10, level-1) n
from    dual
connect by level<=7
),
t2 (comb, lv) as (
select  n, 1
from    t
union all
select  t.n+t2.comb, t2.lv+1
from    t, t2
where   instr(t.n+t2.comb, '3')=0
and     t2.lv<=2
)
,t3 as (
select  comb, count(*) cnt
from    t2
where   lv=3
group by comb
)
,t4 AS (
SELECT a.comb+b.comb comb,sum(a.cnt*b.cnt) as cnt
  from t3 a,t3 b
where  translate(a.comb+b.comb, 'x012', 'x') is null
group by a.comb+b.comb
)
select  TO_CHAR(power(7, 12)-sum(a.cnt*b.cnt))
from    t4 a, t4 b
where   translate(a.comb+b.comb, 'x012', 'x') is null;




oracle sql programming art.sql

16.49 KB, 下載次數: 93

打賞鼓勵一下!

5人打賞

論壇徽章:
527
奧運會紀念徽章:壘球
日期:2008-09-15 01:28:12生肖徽章2007版:雞
日期:2008-11-17 23:40:58生肖徽章2007版:馬
日期:2008-11-18 05:09:48數據庫板塊每日發貼之星
日期:2008-11-29 01:01:02數據庫板塊每日發貼之星
日期:2008-12-05 01:01:03生肖徽章2007版:虎
日期:2008-12-10 07:47:462009新春紀念徽章
日期:2009-01-04 14:52:28數據庫板塊每日發貼之星
日期:2009-02-08 01:01:03生肖徽章2007版:蛇
日期:2009-03-09 22:18:532009日食紀念
日期:2009-07-22 09:30:00
2#
發表于 2016-9-1 00:31 | 只看該作者
打賞按鈕怎么看不到了?要是我有錢就給你五毛。

使用道具 舉報

回復
論壇徽章:
79
生肖徽章2007版:牛
日期:2012-08-02 22:43:00紫蛋頭
日期:2012-12-08 09:43:38鮮花蛋
日期:2012-11-17 12:02:07鮮花蛋
日期:2013-02-05 21:53:34復活蛋
日期:2012-11-17 12:02:07SQL極客
日期:2013-12-09 14:13:35SQL數據庫編程大師
日期:2013-12-06 13:59:43SQL大賽參與紀念
日期:2013-12-06 14:10:50ITPUB季度 技術新星
日期:2012-11-27 10:16:10最佳人氣徽章
日期:2013-03-19 17:24:25
3#
 樓主| 發表于 2016-9-1 08:00 | 只看該作者
newkid 發表于 2016-9-1 00:31
打賞按鈕怎么看不到了?要是我有錢就給你五毛。

我靠,這功能有bug,我發新帖的時候忘了選開啟打賞,等再編輯選開啟,保存不上,我去@喵喵。
5毛?你也做得出

使用道具 舉報

回復
論壇徽章:
403
紫蛋頭
日期:2012-05-21 10:19:41迷宮蛋
日期:2012-06-06 16:02:49奧運會紀念徽章:足球
日期:2012-06-29 15:30:06奧運會紀念徽章:排球
日期:2012-07-10 21:24:24鮮花蛋
日期:2012-07-16 15:24:59奧運會紀念徽章:拳擊
日期:2012-08-07 10:54:50奧運會紀念徽章:羽毛球
日期:2012-08-21 15:55:33奧運會紀念徽章:蹦床
日期:2012-08-21 21:09:51奧運會紀念徽章:籃球
日期:2012-08-24 10:29:11奧運會紀念徽章:體操
日期:2012-09-07 16:40:00
4#
發表于 2016-9-1 08:12 | 只看該作者
udfrog 發表于 2016-9-1 08:00
我靠,這功能有bug,我發新帖的時候忘了選開啟打賞,等再編輯選開啟,保存不上,我去@喵喵。
5毛?你也 ...

這可是newkid的5毛,不能和一般人的比
就像波爾特給你的0.01秒

使用道具 舉報

回復
論壇徽章:
22
林肯
日期:2013-10-25 08:53:562014年世界杯參賽球隊: 澳大利亞
日期:2014-05-26 09:13:482014年世界杯參賽球隊: 阿爾及利亞
日期:2014-05-28 10:46:58喜羊羊
日期:2015-03-04 14:54:422015年新春福章
日期:2015-03-06 11:59:47懶羊羊
日期:2015-06-16 10:25:42雙魚座
日期:2015-07-30 08:48:43白羊座
日期:2015-08-05 20:05:55天蝎座
日期:2015-08-07 22:32:03馬上有車
日期:2014-05-07 11:14:31
5#
發表于 2016-9-1 08:35 | 只看該作者
我原來想賞的,一看沒有那個紅色的打賞按鈕,那就算了吧

使用道具 舉報

回復
論壇徽章:
535
生肖徽章2007版:猴
日期:2008-05-16 11:28:59生肖徽章2007版:馬
日期:2008-10-08 17:01:01SQL大賽參與紀念
日期:2011-04-13 12:08:17授權會員
日期:2011-06-17 16:14:53ITPUB元老
日期:2011-06-21 11:47:01ITPUB官方微博粉絲徽章
日期:2011-07-01 09:45:27ITPUB十周年紀念徽章
日期:2011-09-27 16:30:472012新春紀念徽章
日期:2012-01-04 11:51:22海藍寶石
日期:2012-02-20 19:24:27鐵扇公主
日期:2012-02-21 15:03:13
6#
發表于 2016-9-1 09:29 | 只看該作者
我覺得 newkid 大師的也應該像LZ一樣整理一份PDF(就這個論壇中解過的題)!

使用道具 舉報

回復
認證徽章
論壇徽章:
169
SQL數據庫編程大師
日期:2016-01-13 10:30:43SQL極客
日期:2013-12-09 14:13:35SQL大賽參與紀念
日期:2013-12-06 14:03:45最佳人氣徽章
日期:2015-03-19 09:44:03現任管理團隊成員
日期:2015-08-26 02:10:00秀才
日期:2015-07-28 09:12:12舉人
日期:2015-07-13 15:30:15進士
日期:2015-07-28 09:12:58探花
日期:2015-07-28 09:12:58榜眼
日期:2015-08-18 09:48:03
7#
發表于 2016-9-1 10:59 | 只看該作者

打賞按鈕怎么看不到了?要是我有錢就給你五毛。

使用道具 舉報

回復
論壇徽章:
79
生肖徽章2007版:牛
日期:2012-08-02 22:43:00紫蛋頭
日期:2012-12-08 09:43:38鮮花蛋
日期:2012-11-17 12:02:07鮮花蛋
日期:2013-02-05 21:53:34復活蛋
日期:2012-11-17 12:02:07SQL極客
日期:2013-12-09 14:13:35SQL數據庫編程大師
日期:2013-12-06 13:59:43SQL大賽參與紀念
日期:2013-12-06 14:10:50ITPUB季度 技術新星
日期:2012-11-27 10:16:10最佳人氣徽章
日期:2013-03-19 17:24:25
8#
 樓主| 發表于 2016-9-1 11:04 | 只看該作者
Naldonado 發表于 2016-9-1 10:59
打賞按鈕怎么看不到了?要是我有錢就給你五毛。

凸(艸皿艸 )

使用道具 舉報

回復
認證徽章
論壇徽章:
169
SQL數據庫編程大師
日期:2016-01-13 10:30:43SQL極客
日期:2013-12-09 14:13:35SQL大賽參與紀念
日期:2013-12-06 14:03:45最佳人氣徽章
日期:2015-03-19 09:44:03現任管理團隊成員
日期:2015-08-26 02:10:00秀才
日期:2015-07-28 09:12:12舉人
日期:2015-07-13 15:30:15進士
日期:2015-07-28 09:12:58探花
日期:2015-07-28 09:12:58榜眼
日期:2015-08-18 09:48:03
9#
發表于 2016-9-1 11:09 | 只看該作者
〇〇 發表于 2016-9-1 08:12
這可是newkid的5毛,不能和一般人的比
就像波爾特給你的0.01秒

0.01s夠青蛙生一堆蝌蚪。。

使用道具 舉報

回復
論壇徽章:
403
紫蛋頭
日期:2012-05-21 10:19:41迷宮蛋
日期:2012-06-06 16:02:49奧運會紀念徽章:足球
日期:2012-06-29 15:30:06奧運會紀念徽章:排球
日期:2012-07-10 21:24:24鮮花蛋
日期:2012-07-16 15:24:59奧運會紀念徽章:拳擊
日期:2012-08-07 10:54:50奧運會紀念徽章:羽毛球
日期:2012-08-21 15:55:33奧運會紀念徽章:蹦床
日期:2012-08-21 21:09:51奧運會紀念徽章:籃球
日期:2012-08-24 10:29:11奧運會紀念徽章:體操
日期:2012-09-07 16:40:00
10#
發表于 2016-9-1 13:26 | 只看該作者
solomon_007 發表于 2016-9-1 09:29
我覺得 newkid 大師的也應該像LZ一樣整理一份PDF(就這個論壇中解過的題)!

應該出一本書《蘇神記》

使用道具 舉報

回復

您需要登錄后才可以回帖 登錄 | 注冊

本版積分規則 發表回復

DTCC2020中國數據庫技術大會 限時8.5折

【架構革新 高效可控】2020年9月21日~23日第十一屆中國數據庫技術大會將在北京隆重召開。

大會設置2大主會場,20+技術專場,將邀請超百位行業專家,重點圍繞數據架構、AI與大數據、傳統企業數據庫實踐和國產開源數據庫等內容展開分享和探討,為廣大數據領域從業人士提供一場年度盛會和交流平臺。

http://dtcc.it168.com


大會官網>>
30岁的男人干啥赚钱快赚钱多 股票融资账户怎样操作 二分彩是哪里的 过往开奖记录 江苏快三选号技巧集锦 好玩棋牌游戏平台 时时乐上海开奖结果 最准确的平码公式算法 贵阳捉鸡麻将安卓版 内蒙古十一选五开奖结果 黑龙江p62开奖结果今天晚上
TOP技術積分榜 社區積分榜 徽章 團隊 統計 知識索引樹 積分競拍 文本模式 幫助
  ITPUB首頁 | ITPUB論壇 | 數據庫技術 | 企業信息化 | 開發技術 | 微軟技術 | 軟件工程與項目管理 | IBM技術園地 | 行業縱向討論 | IT招聘 | IT文檔
  ChinaUnix | ChinaUnix博客 | ChinaUnix論壇
CopyRight 1999-2011 itpub.net All Right Reserved. 北京盛拓優訊信息技術有限公司版權所有 聯系我們 
京ICP備09055130號-4  北京市公安局海淀分局網監中心備案編號:11010802021510 廣播電視節目制作經營許可證:編號(京)字第1149號
  
快速回復 返回頂部 返回列表
股票融资账户怎样操作 二分彩是哪里的 过往开奖记录 江苏快三选号技巧集锦 好玩棋牌游戏平台 时时乐上海开奖结果 最准确的平码公式算法 贵阳捉鸡麻将安卓版 内蒙古十一选五开奖结果 黑龙江p62开奖结果今天晚上